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The area enclosed by the curves y=sinx+c...

The area enclosed by the curves `y=sinx+cosx and y=|cosx−sinx|` over the interval `[0,pi/2]` is (a) `4(sqrt2-1)` (b) `2sqrt2(sqrt2-1)` (c) `2(sqrt2+1)` (d) `2sqrt2(sqrt2+1)`

A

`4(sqrt(2)-1)`

B

`2sqrt(2)(sqrt(2)-1)`

C

`2(sqrt(2)+1)`

D

`2sqrt(2)(sqrt(2)+1)`

Text Solution

Verified by Experts

The correct Answer is:
B
Since `sin x and cos x gt 0" for "x in [0,pi//2],` the graph of y= sin x+cos x always lies above the graph of `y=|cos x - sin x|`
Also `cos x gt sin x" for " x in [0,pi//4] and sin gt cos" for " x in [pi//4,pi//2]`
`rArr" Area "=overset(pi//4)underset(0)int((sin x + cos x)-(cos x - sin x))dx+overset(pi//2)underset(pi//4)int((sin x + cos x)-(sin x - cos x))dx`
`=4-2sqrt(2)`
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